AS June 2020 Paper 1 Q17

17 The polar equation of the circle \(C\) is

\[r = a(\cos\theta + \sin\theta)\]

Find, in terms of \(a\), the radius of \(C\).

Fully justify your answer. [4 marks]

Mark scheme

Scheme	Marks	AO
Selects a method to transform the given equation of \(C\) into a standard polar form or Cartesian form, e.g. uses \(r^2 = x^2 + y^2\) and \(x = r\cos\theta\) and \(y = r\sin\theta\); or writes \(r\) in the form \(R(\cos A\cos B + \sin A\sin B)\).	M1	3.1a
Obtains a correct equation in terms of \(x\) and \(y\) only or obtains \(r = a\sqrt{2}\cos\left(\theta - \frac{\pi}{4}\right)\).	A1	1.1b
Correctly completes the square of their quadratic expression or states that the circle must pass through \(O\), and that the maximum value of \(\cos\left(\theta - \frac{\pi}{4}\right)\) is 1.	M1	1.1a
Obtains the correct radius = \(\frac{a}{\sqrt{2}}\).	A1	3.2a
	(4 marks)

Typical solution

\[r^2 = a(r\cos\theta + r\sin\theta)\]\[x^2 + y^2 = a(x + y)\]\[x^2 - ax + y^2 - ay = 0\]\[x^2 - ax + \frac{a^2}{4} + y^2 - ay + \frac{a^2}{4} = \frac{a^2}{2}\]\[\left(x - \tfrac{a}{2}\right)^2 + \left(y - \tfrac{a}{2}\right)^2 = \left(\tfrac{a}{\sqrt{2}}\right)^2\]\[\text{radius} = \frac{a}{\sqrt{2}}\]